logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

Find the equation of normal to the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ at $(x_1,y_1)$ ?

$\begin{array}{1 1}(a)\;\large\frac{b^2y}{y_1}+\frac{a^2x}{x_1}\normalsize = a^2+b^2\\(b)\;\large\frac{b^2y}{y_1}-\frac{a^2x}{x_1}\normalsize = a^2+b^2\\(c)\;\large\frac{b^2y}{y_1}+\frac{a^2x}{x_1}\normalsize = a^2-b^2\\(d)\;\large\frac{b^2y}{x_1}+\frac{a^2x}{y_1}\normalsize = a^2+b^2\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Slope of tangent $=\large\frac{x_1b^2}{y_1a^2}$ (at $(x_1,y_1)$)
Now slope of normal =$-\large\frac{y_1a^2}{x_1b^2}$
Hence equation of normal at $(x_1,y_1)$ is
$(y-y_1)=-\large\frac{y_1a^2}{x_1b^2}$$(x-x_1)$
$b^2(\large\frac{y}{y_1}$$-1)=a^2(1-\large\frac{x}{x_1})$
$\large\frac{b^2y}{y_1}+\frac{a^2x}{x_1}\normalsize = a^2+b^2$
Hence (a) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...